Question

A surveyor wants to find the distance from points A and B to an inaccessible point C. These three points form a triangle. Because point C can be sighted from both A and B, he knows that the measure of < A= 53 degrees and the measure of < B = 61 degrees. In addition, the distance from A to B is 142 meters. Find AC and BC. Draw a diagram.

Answer 1

Answer:

Part a) $AC=135.95\ m$

Part b) $BC=124.14\ m$

The diagram in the attached figure

Step-by-step explanation:

step 1

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$53^o+61^o+m\angle C=180^o$

$114^o+m\angle C=180^o$

$m\angle C=180^o-114^o$

$m\angle C=66^o$

step 2

Find the distance AC

Applying the law of sines

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

see the attached figure to better understand the problem

substitute the given values

$\frac{142}{sin(66^o)}=\frac{AC}{sin(61^o)}$

$AC=\frac{142}{sin(66^o)}(sin(61^o))$

$AC=135.95\ m$ ---> rounded to the nearest hundredth

step 3

Find the distance BC

Applying the law of sines

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

substitute the given values

$\frac{142}{sin(66^o)}=\frac{BC}{sin(53^o)}$

$BC=\frac{142}{sin(66^o)}(sin(53^o))$

$BC=124.14\ m$ ---> rounded to the nearest hundredth